The scalar "[itex]\lambda[/itex]" is an "eigenvalue" for matrix A if and only if there exist a non-zero vector, v, such that [itex]Av= \lambda v[/itex]. It can be shown that [itex]\lambda[/itex] is an eigenvalue for A if and only if it satisfies A's characteristic equation. A vector v, satisfying [itex]Av= \lambda v[/itex] is an eigenvector for A corresponding to eigenvalue [itex]\lambda[/itex] (some people require that v be non-zero to be an "eigenvector" but I prefer to include the 0 vector as an eigenvector for every eigenvalue).

Further, if we can find n independent eigenvectors for A (always true if A has n distinct eigenvalues but often true even if the eigenvalues are not all distinct) then the matrix, P, having those eigenvectors as columns is invertible and [itex]P^{-1}AP=D[/itex] where D is the diagonal matrix having the eigenvalues of A on its diagonal. Then it is also true that [itex]PDP^{-1}= A[/itex] and [tex]A^n= (PDP^{-1})^n= (PDP^{-1})(PDP^{-1})\cdot\cdot\cdot(PDP^{-1})= PD(P^{-1}P)(D)(P^{-1}P)\cdot\cdot\cdot(P^{-1}P)DP= PD^nP^{-1}[/tex]

Of course, [itex]D^n[/itex] is easy to calculate- it is the diagonal matrix having the nth power of the entries in D on its diagonal.

Notice that this is "if we can find n independent eigenvectors for A". (Such a matrix is said to be "diagonalizable" matrix.) There exist non-diagonalizable matrices. They can be put in what is called "Jordan normal form" which is slightly more complicated than a diagonal matrix and it is a little more complicated to find powers.

What HallsofIvy said is true, but I think it's not quite the point of the question.

The Cayley-Hamilton theorem says that the matrix ##A## satisfies its own characteristic equation. For an ##n \times n## matrix, the characteristic equation is of order ##n##, so ##A^n## is a linear combination of ##I, A, \dots, A^{n-1}##. It follows that every power ##A^k## where ##k > n## is also a linear combination of the first ##n-1## powers.

For A ##2 \times 2## matrix, that means every power of ##A## is a linear combination of ##A## and ##I##. That is true even if the eigenvectors are not independent, and you can't diagonalize ##A##.